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Let $m \in \mathbb N$. Find a necessary and sufficient condition for $m$ such that the ideal $(m,x^2+y^2)$ is prime in $\mathbb Z[x,y]$.

I have to find for which $m$ the quotient ring is an integral domain. I do not know how to use the isomorphism theorem: is it ture that $$ \mathbb Z[x,y] /(m,x^2+y^2) \cong (\mathbb Z_m[x])[y]/(x^2+y^2)? $$

What should we do? Thanks in advance.

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1 Answer 1

up vote 4 down vote accepted

Clearly $m = p$ has to be a prime, otherwise its factors are zero divisors.

Also, $\mathbb{Z}[x,y]/(p,x^2+y^2) \cong (\mathbb{Z}_p[x])[y] / (x^2 + y^2)$ is isomorphic by the third isomorphism theorem.

With $\mathbb{Z}_p$ being a field, $\mathbb{Z}_p[x]$ is a PID and so $(x^2 + y^2)$ is a prime ideal in $\mathbb{Z}_p[x,y]$ if and only if $f(y) := x^2 + y^2$ is irreducible over $\mathbb{Z}_p[x]$ (by Gauss's lemma).

$f(y)$ is irreducible if and only if it has no zeros (being a quadratic polynomial), and any zero would have to be of the form $\sqrt{-1}x$; so $f$ is irred. if and only if $-1$ is not a square ($\bmod p$).

By quadratic reciprocity (or rather its supplement) this is if and only if $p \equiv 3 (\bmod 4)$.

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Very nice and elegant. Thanks a lot. –  Romeo Aug 23 '12 at 9:54
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